Question: What is the value of $\dfrac{d}{dx}\left(-\dfrac{1}{x^4}-\dfrac{5}{x^2}\right)$ at $x=2$ ?
The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=2$. Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}-\dfrac{1}{x^4}-\dfrac{5}{x^2} \\\\ &=-x^{-4}-5x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-x^{-4}-5x^{-2} ) \\\\ &=-1\dfrac{d}{dx}(x^{-4})-5\dfrac{d}{dx}(x^{-2}) \\\\ &=-1(-4x^{-5})-5(-2)x^{-3} \\\\ &=4x^{-5}+10x^{-3} \\\\ &=\dfrac{4}{x^5}+\dfrac{10}{x^3} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(-\dfrac{1}{x^4}-\dfrac{5}{x^2}\right)=\dfrac{4}{x^5}+\dfrac{10}{x^3}$. Let's evaluate it at $x=2$ : $\begin{aligned} &\phantom{=}\dfrac{4}{x^5}+\dfrac{10}{x^3} \\\\ &=\dfrac{4}{(2)^5}+\dfrac{10}{(2)^3} \\\\ &=\dfrac{4}{32}+\dfrac{10}{8} \\\\ &=\dfrac{1}{8}+\dfrac{10}{8} \\\\ &=\dfrac{11}{8} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(-\dfrac{1}{x^4}-\dfrac{5}{x^2}\right)$ at $x=2$ is $\dfrac{11}{8}$.